144 5.6 Nondifferentiable Convex Functions and Subdifferentials Thus, u≤ lim x→x+ 0 f (x)−f (x0) x−x0 =f ′+(x0). Similarly, we have u·(x−x0) ≤f (x)−f (x0) for all x <x0. Thus, u≥ f (x)−f (x0) x−x0 for all x <x0. This implies u≥f ′ − (x0). So ∂ f (x0) ⊂[ f ′ − (x0), f ′+(x0)]. To prove the reverse inclusion, take u∈[ f ′ − (x0), f ′+(x0)]. By Theorem 5.6.2 sup x<x0 φx0(x)=f ′ − (x0) ≤u≤f ′+(x0)= inf x>x0 φx0(x). Using the upper estimate by f ′+(x0) for u, one has u≤φx0(x)= f (x)−f (x0) x−x0 for all x >x0. It follows that u·(x−x0) ≤f (x)−f (x0) for all x ≥x0. Similarly, one also has u·(x−x0) ≤f (x)−f (x0) for all x <x0. Thus, (5.8) holds and, hence, u∈∂ f (x0). Therefore, (5.10) holds. □ Corollary 5.6.4 Let f : R→Rbe a convex function and let x0 ∈R. Then f is differentiable at x0 if and only if ∂ f (x0) is a singleton. In this case, ∂ f (x0)={f ′(x0)}. Proof: Suppose f is differentiable at x0. Then f ′ − (x0)=f ′+(x0)=f ′(x0). By Theorem 5.6.3, ∂ f (x0)=[ f ′ − (x0), f ′+(x0)]={f ′(x0)}. Thus, ∂ f (x0) is a singleton. Conversely, if ∂ f (x0) is a singleton, we must have f ′ − (x0)=f ′+(x0). Thus, f is differentiable at x0. □
RkJQdWJsaXNoZXIy NTc4NTAz